Problem: Perform the row operation, $-5R_1\rightarrow R_1$, on the following matrix. $\left[\begin{array} {ccc} -9 & -2 & 3 & 7 \\ 1 & -8 & 4 & 5 \\ 2 & -4 & 2 & -6 \end{array} \right] $
Explanation: Background There are three basic row operations that can be performed on matrices. $R_i \leftrightarrow R_j$. This symbol tells us to interchange rows $i$ and $j$. $cR_i \rightarrow R_i$. This symbol tells us to multiply a row $i$ by a constant $c$. $R_i + cR_j \rightarrow R_i$. This symbol tells us to add $c$ times row $j$ to row $i$. Finding the new row to be used For the given matrix, $R_1$ is given below. $R_1=\left[\begin{array} {ccc} -9 & -2 & 3 & 7 \end{array} \right]$ We are asked to perform the row operation, $-5R_1\rightarrow R_1$. Therefore, we must multiply $R_1$ by $-5$. $\begin{aligned}-5R_1 &= -5\left[\begin{array} {ccc} -9 & -2 & 3 & 7 \end{array} \right] \\\\&=\left[\begin{array} {ccc} 45 & 10 & -15 & -35 \end{array} \right]\end{aligned}$ Substituting the row Now, we must substitute row $R_1$ with $-5R_1$. $\left[\begin{array} {ccc} {-9} & {-2} & {3} & {7} \\ 1 & -8 & 4 & 5 \\ 2 & -4 & 2 & -6 \end{array} \right]\xrightarrow{-5R_1\rightarrow R_1}\left[\begin{array} {ccc} {45} & {10} & {-15} & {-35} \\ 1 & -8 & 4 & 5 \\ 2 & -4 & 2 & -6 \end{array} \right]$ Summary Our resultant matrix is the following. $\left[\begin{array} {ccc} 45 & 10 & -15 & -35 \\ 1 & -8 & 4 & 5 \\ 2 & -4 & 2 & -6 \end{array} \right]$